3.100 \(\int \frac {1}{(c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=39 \[ \frac {2 x}{3 c^2 \sqrt {c+d x^2}}+\frac {x}{3 c \left (c+d x^2\right )^{3/2}} \]

[Out]

1/3*x/c/(d*x^2+c)^(3/2)+2/3*x/c^2/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {192, 191} \[ \frac {2 x}{3 c^2 \sqrt {c+d x^2}}+\frac {x}{3 c \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(-5/2),x]

[Out]

x/(3*c*(c + d*x^2)^(3/2)) + (2*x)/(3*c^2*Sqrt[c + d*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx &=\frac {x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c}\\ &=\frac {x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x}{3 c^2 \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.74 \[ \frac {x \left (3 c+2 d x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(-5/2),x]

[Out]

(x*(3*c + 2*d*x^2))/(3*c^2*(c + d*x^2)^(3/2))

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fricas [A]  time = 0.62, size = 47, normalized size = 1.21 \[ \frac {{\left (2 \, d x^{3} + 3 \, c x\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{2} d^{2} x^{4} + 2 \, c^{3} d x^{2} + c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*d*x^3 + 3*c*x)*sqrt(d*x^2 + c)/(c^2*d^2*x^4 + 2*c^3*d*x^2 + c^4)

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giac [A]  time = 0.60, size = 27, normalized size = 0.69 \[ \frac {x {\left (\frac {2 \, d x^{2}}{c^{2}} + \frac {3}{c}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*x*(2*d*x^2/c^2 + 3/c)/(d*x^2 + c)^(3/2)

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maple [A]  time = 0.00, size = 26, normalized size = 0.67 \[ \frac {\left (2 d \,x^{2}+3 c \right ) x}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x^2+c)^(5/2),x)

[Out]

1/3*x*(2*d*x^2+3*c)/(d*x^2+c)^(3/2)/c^2

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maxima [A]  time = 1.34, size = 31, normalized size = 0.79 \[ \frac {2 \, x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

2/3*x/(sqrt(d*x^2 + c)*c^2) + 1/3*x/((d*x^2 + c)^(3/2)*c)

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mupad [B]  time = 4.79, size = 28, normalized size = 0.72 \[ \frac {2\,x\,\left (d\,x^2+c\right )+c\,x}{3\,c^2\,{\left (d\,x^2+c\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c + d*x^2)^(5/2),x)

[Out]

(2*x*(c + d*x^2) + c*x)/(3*c^2*(c + d*x^2)^(3/2))

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sympy [B]  time = 0.83, size = 95, normalized size = 2.44 \[ \frac {3 c x}{3 c^{\frac {7}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 3 c^{\frac {5}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {2 d x^{3}}{3 c^{\frac {7}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 3 c^{\frac {5}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x**2+c)**(5/2),x)

[Out]

3*c*x/(3*c**(7/2)*sqrt(1 + d*x**2/c) + 3*c**(5/2)*d*x**2*sqrt(1 + d*x**2/c)) + 2*d*x**3/(3*c**(7/2)*sqrt(1 + d
*x**2/c) + 3*c**(5/2)*d*x**2*sqrt(1 + d*x**2/c))

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